M1 Question

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  1. NutterFrutter's Avatar
    1 05-05-2012  10:50
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    M1 Question
    I'm really struggling to understand question 3a on the Solomon Paper L. I've looked at the markscheme but still can't understand how to get to the answer.
    Could someone please explain the question and the markscheme?
    Attached Files
  2. File Type: pdf M1 Solomon Press Paper L.pdf (142.4 KB, 42 views)
  3. File Type: pdf M1 Solomon Press Paper L MS.pdf (139.6 KB, 177 views)
  4. raheem94's Avatar
    2 05-05-2012  11:10
    • TSR Demigod
    • Posts: 5,512
    Re: M1 Question
    (Original post by NutterFrutter)
    I'm really struggling to understand question 3a on the Solomon Paper L. I've looked at the markscheme but still can't understand how to get to the answer.
    Could someone please explain the question and the markscheme?
    They are just resolving the forces both vertically and horizontally.




    First resolve horizontally, this will give, T_Bcos(90- \alpha) = T_Acos\alpha

    Remember, cos(90- \alpha) =sin\alpha

    So the equation becomes \displaystyle T_Bsin\alpha = T_Acos\alpha \implies T_B \times \frac35 = T_A \times \frac45 \implies T_A=\frac{3T_B}{4}

    Now resolve vertically,
    T_Asin\alpha + T_Bsin(90-\alpha) = 1000g

    Remember, sin(90- \alpha) =cos\alpha

    So the equation becomes, \displaystyle T_Asin\alpha + T_Bcos\alpha = 1000g

    Now i think you can do it further.

    Do you get it?
  5. NutterFrutter's Avatar
    3 05-05-2012  11:16
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    Re: M1 Question
    (Original post by raheem94)
    They are just resolving the forces both vertically and horizontally.




    First resolve horizontally, this will give, T_Bcos(90- \alpha) = T_Acos\alpha

    Remember, cos(90- \alpha) =sin\alpha

    So the equation becomes \displaystyle T_Bsin\alpha = T_Acos\alpha \implies T_B \times \frac35 = T_A \times \frac45 \implies T_A=\frac{3T_B}{4}

    Now resolve vertically,
    T_Asin\alpha + T_Bsin(90-\alpha) = 1000g

    Remember, sin(90- \alpha) =cos\alpha

    So the equation becomes, \displaystyle T_Asin\alpha + T_Bcos\alpha = 1000g

    Now i think you can do it further.

    Do you get it?
    Ahhh, I didn't think C2 methods would be required for a M1 paper as the only prerequisite is C1, but it is a Solomon Press paper. :lol: Anyway, thanks for the help, I understand it now.
  6. raheem94's Avatar
    4 05-05-2012  11:19
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    Re: M1 Question
    (Original post by NutterFrutter)
    Ahhh, I didn't think C2 methods would be required for a M1 paper as the only prerequisite is C1, but it is a Solomon Press paper. :lol: Anyway, thanks for the help, I understand it now.
    I actually would have done this question using a vector triangle, rather than the mark scheme technique.
  7. NutterFrutter's Avatar
    5 05-05-2012  11:20
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    Re: M1 Question
    (Original post by raheem94)
    I actually would have done this question using a vector triangle, rather than the mark scheme technique.
    Could you explain how to do it using a vector triangle? :unsure:
  8. raheem94's Avatar
    6 05-05-2012  11:37
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    • Posts: 5,512
    Re: M1 Question
    (Original post by NutterFrutter)
    Could you explain how to do it using a vector triangle? :unsure:
    See the image below, it is the vector triangle for it,




    Now to find T_B , we will do,
    \displaystyle sin(90-\alpha) = \frac{T_B}{1000g} \implies cos\alpha = \frac{T_B}{1000g} \implies \frac45 \times 1000g = T_B \\ \implies \boxed{T_B=7840N}

    Do i make any sense?
    Last edited by raheem94; 05-05-2012 at 11:38.
  9. NutterFrutter's Avatar
    7 05-05-2012  11:45
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    • Posts: 4,631
    Re: M1 Question
    (Original post by raheem94)
    See the image below, it is the vector triangle for it,




    Now to find T_B , we will do,
    \displaystyle sin(90-\alpha) = \frac{T_B}{1000g} \implies cos\alpha = \frac{T_B}{1000g} \implies \frac45 \times 1000g = T_B \\ \implies \boxed{T_B=7840N}

    Do i make any sense?
    Yes, it makes sense. Thanks.
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